Proof verification: Proving or disproving that $(\Bbb R^+, \oplus, \otimes)$ as defined is a field

Question

The problem I have been working on, in summary:

Let $R$ denote the set of positive real numbers. Define addition, denoted $\oplus$, and multiplication, dentoed $\otimes$, respectively by $a \oplus b = ab$ and $a \otimes b = a^{\log(b)}$. Prove or disprove that $(R,\oplus,\otimes)$ is a field.

So far, I have been working on proving the following axioms, and would like to check the validity of my solutions:

• $(1) \; a \oplus b = ab =ba =b \oplus a$, thus true
• $(2) \; a \oplus (b \oplus c) = (a \oplus b) \oplus c = (ab)c = a(bc)$, thus true
• $(3) \; 0 \oplus a = a \oplus 0 = a; a(0) = 0$, thus false
• $(4) \; a \oplus (-a)=a(-a)=0; a(-a)=-(a^2)$, thus false

Can anyone confirm that this is right? Or if it's wrong, please tell me what I'm doing wrong?

Answer 1

The additive identity ("0") here may not be equal to the real number 0. In fact, the number 0 isn't even in the base set. You need to check whether some element can serve as the identity under $\oplus$. For instance, here the real number 1 is the additive identity. Similarly, the additive inverse, multiplicative identity and multiplicative inverse may also be different.

Answer 2

Your title is misleading: the question is asking whether $(R,\oplus,\otimes)$ is a field, not $(R,+,*)$. Of course, $R$ is pretty bad notation too.

For (3) I presume you want to check that $R$ has an additive identity. It certainly isn't $0$ as $0\notin R$. We want an element $e$ with $e\oplus a=a$ for all $a\in R$, that is $ea=a$ for all $a>0$. Thus $e=1$ is the additive identity.

Likewise in (4) you need to prove for all $a\in R$, there is $b\in R$ with $a\oplus b=1$, that is $ab=1$, so $b=1/a$ etc.

Answer 3

Your results are not quite correct. The first two are fine, but the latter two have problems resulting from what I suspect to be confusions about the notation.

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In this context, $0$ generally does not denote the number zero, but rather the additive identity of the group you're working with. Sometimes I see this denoted $e$ to avoid that painful ambiguity, and always denote it as such in my own work.

Then, in $(3)$, what you're seeking to prove is that, for all $a \in \Bbb R^+$,

$$a \oplus e = e \oplus a = a$$

for whatever this identity $e$ is. It would be prudent to first find that $e$. Since $\oplus$ is defined by real multiplication, then, you seek $e$ such that

$$a \oplus e = e \oplus a = a \iff a \cdot e = e \cdot a = a$$

Clearly, $e = 1$. With that in mind, showing this axiom's truth should not be difficult.

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Similarly, let's drop the $0$ notation for $(4)$. Let $e$ be the additive identity of the group $(\Bbb R^+, \oplus)$. Then we want to show the inverse axiom: for all $a \in \Bbb R^+$, there exists an inverse element $(-a) \in \mathbb R^+$ such that

$$a \oplus (-a) = (-a) \oplus a = e \iff a \cdot (-a) = (-a) \cdot (a) = 1$$

(The right side of the $\iff$ comes from applying the definition of $\oplus$ and what we already found $e$ to be in the previous section.)

Another notational thing you seem to be (understandably) confused about: $(-a)$ denotes the "additive inverse," i.e. the inverse with respect to this operation of addition. It's, again, a somewhat ambiguous notation, and does not mean $-1 \times a$. Think of it just like another number. For example, you could reframe this axiom as:

"For each $a$ in our group $(G,\ast)$, there exists a $b$ in our group such that $$a \ast b = b \ast a = e$$ where $e$ is the identity of the group."

Keep in mind what you know about multiplicative inverses over the real numbers to prove this axiom (since we define our addition operation $\oplus$ in your problem by that kind of multiplication).

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Some further notes:

It is important to know where your results come from in this context. For example, you get $ab=ba$ because of the commutativity of real number multiplication, i.e. from the fact that $(\Bbb R, +. \cdot)$ - with $+,\cdot$ defined as they usually are - is a field. This fact does not come from thin air and can be rigorously justified, and should be - particularly if this is a homework assignment, you'll likely be docked points for not saying where this comes from. (I imagine like most courses it will be assumed that the reals do indeed form a field, i.e. you will not have to prove it's a field, but you can with a sufficient understanding of real analysis I believe.)

It would also be prudent to include a lot more words in general - state the identity you're trying to prove, and why that identity is satisfied or not. Your work can be hard to follow at some points. Explain, don't do everything in equations, just expecting everyone to understand.